--- title: "Hill road earthwork with cross sections" subject: "Estimating and Costing" chapter: 6 key: "semester_7/estimating_and_costing:hill-road-u-cross-sections" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-u-cross-sections site: BCT Engineering --- # Hill road earthwork with cross sections Estimating and Costing · Chapter 6: Detailed Estimate - Asked in: 2066 Bhadra (old course), 6 marks - Video: [Calculation of Volume of Earthwork from Cross Sections | Road Estimate | Earthwork Calculation](https://www.youtube.com/watch?v=zMl6OVLeKWM) — Civil Engineering Edu - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/hill-road-u-cross-sections - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-6 ## Question Find out the quantity of a hill road when the following data are given: formation width is 10 m. Side slope in cutting and filling are (1:1 and 2:1) respectively. | Chainage | Depth of cutting at centre line | Cross slope of ground | |---|---|---| | 0 | 0.5 m | 10:1 | | 30 | 0.30 m | 12:1 | | 60 | 1.00 m | 10:1 | Draw cross section at each point. ## Short answer Using mean area times 30 m: cutting = (5.556+3.362)/2×30 + (3.362+11.364)/2×30 = 354.66 m³; filling = (0+0.098)/2×30 + (0.098+0)/2×30 = 2.94 m³. ## Answer #### Given data and assumptions - Formation width $B = 10$ m; side slope in cutting $1:1$, in filling $2:1$ (H:V); distance between sections 30 m. #### Method Ground has a transverse (cross) slope $n:1$ (H:V), so each section is a side-hill section. With centre height $h$ (cutting +, filling −) above/below formation level, formation width $B$ and side slope $s:1$: $$ \begin{aligned} x_u &= \frac{n\,(B/2 + s h)}{n - s},\quad d_u = h + \frac{x_u}{n}\\ x_d &= \frac{n\,(B/2 + s h)}{n + s},\quad d_d = h - \frac{x_d}{n}\\ A &= \tfrac12 B h + \tfrac12\left[\tfrac{B}{2}(d_u + d_d) + h\,(x_u + x_d - B)\right] \end{aligned} $$ Here $x_u, x_d$ are the horizontal distances of the up-hill and down-hill toes from the centre line and $d_u, d_d$ the heights of the ground above formation at the toes (formulas for cutting). For a section wholly in filling the roles swap: the down-hill toe is the far one, $x_d = n(B/2 + s h)/(n - s)$ and $x_u = n(B/2 + s h)/(n + s)$, with $h$ the filling height and $s$ the filling slope. Where the ground line crosses the formation level inside the road width (part cutting, part filling), the cutting and filling areas are found separately from the co-ordinates of the section (toe points from the same intersection equations). Volume between sections = mean area × distance, taken separately for cutting and filling. #### Sectional areas | Chainage | Centre ht. $h$ (m) | Cutting area (m²) | Filling area (m²) | |---|---:|---:|---:| | 0+000 | cut 0.50 | 5.556 | 0.000 | | 0+030 | cut 0.30 | 3.362 | 0.098 | | 0+060 | cut 1.00 | 11.364 | 0.000 | Working for one section: Take chainage 0+030: centre cutting depth $h = 0.30$ m, $n = 12$, $B = 10.0$ m. - Ground height above formation at the uphill edge $= 0.717$ m, at the downhill edge $= -0.117$ m. - Toe of the side slope on the uphill side at $x = 5.782$ m from the centre line (ground above formation by 0.782 m); downhill toe at $x = 5.280$ m (ground below formation by 0.140 m). - Area from the co-ordinates of the polygon: cutting $= 3.362$ m², filling $= 0.098$ m². #### Cross-sections (vertical scale exaggerated) Chainage 0+000 (centre cutting 0.50 m, cross slope 10:1; area of cutting 5.556 m², filling 0.000 m²) ```text ... ##..... ## ........ # ....... ## ........ # ....... # ........ ########################################## ... ``` (. ground line, # formation and side slopes) Chainage 0+030 (centre cutting 0.30 m, cross slope 12:1; area of cutting 3.362 m², filling 0.098 m²) ```text .. .##..... # ....... # ........ # ........ # ........ ########################################## .....##. .. ``` (. ground line, # formation and side slopes) Chainage 0+060 (centre cutting 1.00 m, cross slope 10:1; area of cutting 11.364 m², filling 0.000 m²) ```text ..#. ##........ ## .......... ## ......... ## ......... ## .......... # ##.. ## ## ###################################### ``` (. ground line, # formation and side slopes) #### Volumes | Between | Distance (m) | Cutting (m³) | Filling (m³) | |---|---:|---:|---:| | 0+000 – 0+030 | 30.0 | 133.76 | 1.47 | | 0+030 – 0+060 | 30.0 | 220.88 | 1.47 | | **Total** | 60.0 | **354.64** | **2.94** | **Answer: Earthwork in cutting = 354.64 m³; earthwork in filling = 2.94 m³.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At 0+030 (cut 0.30 m, cross slope 12:1, B = 10 m) the ground at the down-hill edge relative to formation is: A. +0.717 m B. -0.117 m C. -0.417 m D. +0.117 m Answer: B. Down-hill edge = 0.30 - 5/12 = -0.117 m, below formation. (Key point: Edge height) 2. Why is the up-hill toe (5.782 m at 0+030) further out than the half formation width of 5 m? A. The ground rises on that side so the 1:1 side slope must climb to meet it B. The formation widens in cutting C. The filling slope is used D. Because the drawing is exaggerated Answer: A. Rising ground meets the side slope only after it has gone beyond the formation edge. (Key point: Toe on up-hill side) 3. Filling areas are 0.000 (0+000), 0.098 (0+030), 0.000 (0+060) m^2, 30 m apart. Total filling volume is about: A. 5.88 m^3 B. 2.94 m^3 C. 0.098 m^3 D. 1.47 m^3 Answer: B. Each 30 m reach gives 0.098/2 x 30 = 1.47 m^3; two reaches total 2.94 m^3. (Key point: Filling volume) ## Related in Estimating and Costing - [Hill road earthwork, cross slope 1 in 5 (0-60)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-r-cross-slope-1-5) — Estimating and Costing, chapter 6 - [Hill road earthwork, chainage 0-600 (12 marks)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-l-0-600) — Estimating and Costing, chapter 6 - [Hill road earthwork, chainage 130-280](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-formation-8-table-a) — Estimating and Costing, chapter 6 - [Hill road earthwork, chainage 0-150 (50 m)](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-j-formation-10) — Estimating and Costing, chapter 6 - [Hill road earthwork, chainage 12-15](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-q-chain-12-15) — Estimating and Costing, chapter 6 - [Hill road earthwork, cutting 4 m / banking 6 m](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Ahill-road-k-cut4-fill6) — Estimating and Costing, chapter 6 ## Related in other subjects - [Earthwork volume, formation 1201.85, 1.65 m depth](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-earthwork-formation-1201-85) — Surveying I, chapter 9 - [Cutting volume, 60 m, GL 540.70, formation 538.20](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-540-70) — Surveying I, chapter 9, asked 7× - [Cutting volume, 60 m, GL 1181.50, formation 1179.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-1181-50-formation-1179) — Surveying I, chapter 9, asked 8× - [Cutting volume, 60 m, formation level 279.00](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Avolume-cutting-60m-gl-281-50-formation-279) — Surveying I, chapter 9, asked 7×